2022 AIME II Problems/Problem 10

Problem

Find the remainder when $\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}$ is divided by $1000$ .

Solution

To solve this problem, we need to use the following result:

$\sum_{i=n}^m \binom{i}{k} = \binom{m+1}{k+1} - \binom{n}{k+1} .$

Now, we use this result to solve this problem.

We have $\begin{align*} \sum_{i=3}^{40} \binom{\binom{i}{2}}{2} & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \\ & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} \left( i \left( i - 1 \right) \left( i \left( i - 1 \right) - 2 \right) \right) \\ & = \frac{1}{8} \sum_{i=3}^{40} \left( i \left( i - 1 \right) \left( \left( i - 2 \right) \left( i - 3 \right) + 4 \left( i - 2 \right) \right) \right) \\ & = 3 \left( \sum_{i=3}^{40} \binom{i}{4} + \sum_{i=3}^{40} \binom{i}{3} \right) \\ & = 3 \left( \binom{41}{5} - \binom{3}{5} + \binom{41}{4} - \binom{3}{4} \right) \\ & = 3 \left( \binom{41}{5} + \binom{41}{4} \right) \\ & = 3 \cdot \frac{41 \cdot 40 \cdot 39 \cdot 38}{5!} \left( 37 + 5 \right) \\ & = 3 \cdot 41 \cdot 13 \cdot 38 \cdot 42 \\ & = 38 \cdot 39 \cdot 41 \cdot 42 \\ & = \left( 40 - 2 \right) \left( 40 - 1 \right) \left( 40 + 1 \right) \left( 40 + 2 \right) \\ & = \left( 40^2 - 2^2 \right) \left( 40^2 - 1^2 \right) \\ & = \left( 40^2 - 4 \right) \left( 40^2 - 1 \right) \\ & = 40^4 - 40^2 \cdot 5 + 4 . \end{align*}$

Therefore, modulo 1000, $\sum_{i=3}^{40} \binom{\binom{i}{2}}{2} \equiv \boxed{\textbf{(004) }}$ .

~Steven Chen (www.professorchenedu.com)

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2022 AIME II Problems/Problem 10

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